Google
Câu hỏi: Cho hình thang vuông $ABCD$, đường cao $AB=2a$, $AD=a$, $BC=4a$. Gọi $I$ là trung điểm của $CD$, $J$ là điểm di động trên cạnh $BC$. Tính $BJ$ sao cho $AJ$ và $BI$ vuông góc.
A. $\dfrac{3a}{4}$.
B. $\dfrac{4a}{5}$.
C. $a$.
D. $\dfrac{5a}{6}$.
Biểu diễn $\overrightarrow{BI}$ theo hai vectơ $\overrightarrow{BA}$ và $\overrightarrow{BC}$.
Ta có $\overrightarrow{BI}$ $=\dfrac{1}{2}$ $\left( \overrightarrow{BD}+\overrightarrow{BC} \right)$ $=\dfrac{1}{2}\overrightarrow{BD}+\dfrac{1}{2}\overrightarrow{BC}$ $=\dfrac{1}{2}\left( \overrightarrow{BA}+\overrightarrow{AD} \right)+\dfrac{1}{2}\overrightarrow{BC}$ $=\dfrac{1}{2}\overrightarrow{BA}+\dfrac{1}{2}\overrightarrow{AD}+\dfrac{1}{2}\overrightarrow{BC}$
$=\dfrac{1}{2}\overrightarrow{BA}+\dfrac{1}{2}.\dfrac{1}{4}\overrightarrow{BC}+\dfrac{1}{2}\overrightarrow{BC}$ $=\dfrac{1}{2}\overrightarrow{BA}+\dfrac{5}{8}\overrightarrow{BC}$.
Đặt $\overrightarrow{BJ}=k\overrightarrow{BC}$. Biểu diễn $\overrightarrow{AJ}$ theo hai vectơ $\overrightarrow{BA}$ và $\overrightarrow{BC}$.
Ta có $\overrightarrow{AJ}=\overrightarrow{BJ}-\overrightarrow{BA}$ $=k\overrightarrow{BC}-\overrightarrow{BA}$.
Do $AJ$ và $BI$ vuông góc nên $\overrightarrow{AJ}$. $\overrightarrow{BI}$ $=0$.
Suy ra: $\left( \dfrac{1}{2}\overrightarrow{BA}+\dfrac{5}{8}\overrightarrow{BC} \right)\left( k\overrightarrow{BC}-\overrightarrow{BA} \right)=0$
$\Leftrightarrow \dfrac{k}{2}\overrightarrow{BA}.\overrightarrow{BC}-\dfrac{1}{2}\overrightarrow{BA}.\overrightarrow{BA}+\dfrac{5k}{8}\overrightarrow{BC}.\overrightarrow{BC}-\dfrac{5}{8}\overrightarrow{BC}.\overrightarrow{BA}=0$
$\Leftrightarrow -\dfrac{1}{2}B{{A}^{2}}+\dfrac{5k}{8}B{{C}^{2}}=0$ ( do $BA\bot BC$ nên $\overrightarrow{BA}.\overrightarrow{BC}=0$ )
$\Leftrightarrow -\dfrac{1}{2}.4{{a}^{2}}+\dfrac{5k}{8}.16{{a}^{2}}=0$ $\Leftrightarrow k=\dfrac{1}{5}$.
Vậy $\overrightarrow{BJ}=\dfrac{1}{5}\overrightarrow{BC}$, suy ra $BJ=\dfrac{1}{5}BC=\dfrac{4a}{5}.$
A. $\dfrac{3a}{4}$.
B. $\dfrac{4a}{5}$.
C. $a$.
D. $\dfrac{5a}{6}$.
Biểu diễn $\overrightarrow{BI}$ theo hai vectơ $\overrightarrow{BA}$ và $\overrightarrow{BC}$.
Ta có $\overrightarrow{BI}$ $=\dfrac{1}{2}$ $\left( \overrightarrow{BD}+\overrightarrow{BC} \right)$ $=\dfrac{1}{2}\overrightarrow{BD}+\dfrac{1}{2}\overrightarrow{BC}$ $=\dfrac{1}{2}\left( \overrightarrow{BA}+\overrightarrow{AD} \right)+\dfrac{1}{2}\overrightarrow{BC}$ $=\dfrac{1}{2}\overrightarrow{BA}+\dfrac{1}{2}\overrightarrow{AD}+\dfrac{1}{2}\overrightarrow{BC}$
$=\dfrac{1}{2}\overrightarrow{BA}+\dfrac{1}{2}.\dfrac{1}{4}\overrightarrow{BC}+\dfrac{1}{2}\overrightarrow{BC}$ $=\dfrac{1}{2}\overrightarrow{BA}+\dfrac{5}{8}\overrightarrow{BC}$.
Đặt $\overrightarrow{BJ}=k\overrightarrow{BC}$. Biểu diễn $\overrightarrow{AJ}$ theo hai vectơ $\overrightarrow{BA}$ và $\overrightarrow{BC}$.
Ta có $\overrightarrow{AJ}=\overrightarrow{BJ}-\overrightarrow{BA}$ $=k\overrightarrow{BC}-\overrightarrow{BA}$.
Do $AJ$ và $BI$ vuông góc nên $\overrightarrow{AJ}$. $\overrightarrow{BI}$ $=0$.
Suy ra: $\left( \dfrac{1}{2}\overrightarrow{BA}+\dfrac{5}{8}\overrightarrow{BC} \right)\left( k\overrightarrow{BC}-\overrightarrow{BA} \right)=0$
$\Leftrightarrow \dfrac{k}{2}\overrightarrow{BA}.\overrightarrow{BC}-\dfrac{1}{2}\overrightarrow{BA}.\overrightarrow{BA}+\dfrac{5k}{8}\overrightarrow{BC}.\overrightarrow{BC}-\dfrac{5}{8}\overrightarrow{BC}.\overrightarrow{BA}=0$
$\Leftrightarrow -\dfrac{1}{2}B{{A}^{2}}+\dfrac{5k}{8}B{{C}^{2}}=0$ ( do $BA\bot BC$ nên $\overrightarrow{BA}.\overrightarrow{BC}=0$ )
$\Leftrightarrow -\dfrac{1}{2}.4{{a}^{2}}+\dfrac{5k}{8}.16{{a}^{2}}=0$ $\Leftrightarrow k=\dfrac{1}{5}$.
Vậy $\overrightarrow{BJ}=\dfrac{1}{5}\overrightarrow{BC}$, suy ra $BJ=\dfrac{1}{5}BC=\dfrac{4a}{5}.$
Đáp án B.